When we promote fields to operators, do all fields become the exact same type of operator? For example in quantising the Dirac field (spinor) vs K-G field (scalar), these aren't different types of object once they become operators?

@Charlie what do you mean by "type"?

mmm I don't mean identical relationships with other operators, but they are just "linear operators on the Hilbert space"

which now that I think about it they must be in order for commutation relations to exist between them

@Charlie ...what do you mean by "kind of field"?

:(

just the type of object placed at each point in the field

scalar field, vector field etc

the finite-dimensional representation on the target space of the fields is required to play nice with the infinite-dimensional representation on the Fock space, see e.g. physics.stackexchange.com/a/174908/50583, this is once again the "two representations" issue

if you construct the fields Weinberg-style this is a consequence of their construction, if you don't, it can be an assumption

@Charlie I will always be pedantic and require you to make explicit what property you're thinking about instead of accepting "type" as a sufficient identifier :P

Why do the two representations need to play nice?

I can't think of a situation in which they would interact

And being pedantic is good :p

@Charlie the "play nice" condition means that it doesn't matter if you first apply a Poincaré transform and then quantize or first quantize and then Poincaré transform

I.e. this condition is what says that QFT is really special relativistic because it yields the same quantum theory for all inertial observers

oh ok

Another Player enters the arena!

twitter.com/D_Roopa_IPS/status/1305176827774656512?s=20

@ACuriousMind I didn't mean to sound that way, sorry pal.

Ehrenfest theorem gives us a way to get classical trajectories out of quantum states in the form of expectation values. For certain wave packets, we can, to a good approximation replace the operators with the mean value and this way treat certain systems semi-classically.

Does this mean that the state we are considering doesn’t change much (subject to the precision of measurement) so as to consider the state to be the same wavepacket after measurement even though it isn’t an eigenstate of that operator?

I roughly guess yes.

EFT only gives classical trajectories only in limiting cases i.e. $\hbar \rightarrow 0 \, or \, m \rightarrow + \infty$

Did you know that large and well formed iron pyrite crystals can be bought for only £10:

I am seriously tempted. Not that I have any use for iron pyrite crystals, but they look cool :-)

@JohnRennie What are those? Iron cubes?

Iron pyrite is ferrous sulphide FeS.

sounds cool!

It's called "Fools gold" because it has a shiny appearance just like gold.

Wow! Even, I fell for it.

I don't know what will I do with it, but I need them, I'm getting an urge to take few in palms and squeeze them...

Can they be eaten?

Edible too?

That's exactly what I feel. They just look so amazing.

hehehe :)

I don't think ferrous sulphide is poisonous but I doubt it would taste good :-)

hehehe funny!

See I found a cool puzzle!

@abhas_RewCie Which book is this?

By any chance is this Nob Yoshigahara's book?

@abhas_RewCie 12

@Mayank No idea, probably Nob's.

@SuperfastJellyfish Abxolutterli Correckt!

THINGS THAT HAPPEN IN INDIA ONLY

hehehehehe

@ACuriousMind didn't get any answer. Asked hours ago and it is a simple question with "yes" or "no" answer

@Forex007 Our site does not aim to answer questions in real-time - many users only visit the site once a day or even less often. Also, you asked your question only two hours ago, which barely qualifies for the plural in "hours ago". Please just be a bit more patient.

@PhysicsMeta Let's see how many mods are left after October 7, when SE will try to enforce the so-called "new moderator agreement".

@Charlie A classical vector field transforms as $A' = \Lambda A$ under a Lorentz transformation, since a field becomes an operator when quantised, one has to ask what are the conditions which mean that an operator behaves like a vector under Lorentz transformations, i.e. is a 'vector operator', with scalar, spinor etc... analogues and then one has to verify the fields one quantises do behave that way

In non-relativistic mechanics one encounters this after angular momentum has been defined, talking about 'vector operators' etc

Wait the field operators themselves transform under the Lorentz group?

or the coordinates that label them

The field operators do transform, yes

oh, that is definitely new information for me then

Skim these and these and these slides (taken from here)

Since 1) they are distributions, and therefore depend on coordinates 2) they also are representations of the Poincaré group

ie the operator for the EM potential $A$ gives you a vector in the end

To be precise, $\langle \Psi, \hat{A}[f] \Phi \rangle$ is a vector

Basically because you promote a field to an operator, to do something like a Lorentz transformation you have to basically do a change of basis on the operator i.e. conjugate it $\psi_a \to \psi_a' = U^{\dagger} \psi_a U = S_{ab} \psi_b$ (slide 12 of lecture 2 there) and this can then be equal to different things when the result is a scalar, vector, spinor etc...

@Slereah distributions are just too hard and can basically be ignored, is there a single benefit to caring about them

@bolbteppa The CCR, for a start

Dirac delta "functions"

Can't even bring yourself to pretend that they are functions!

I'm still a bit lost on how we formally do coordinate transformations on a manifold, it makes sense on a vector space where we have objects we can act on with some group representation but we can do this with manifolds

this essentially goes back to how we can formally work with polar coordinates, which aren't well defined on a vector space

Coordinate transforms on a manifold are basically that, locally, for a point of your manifold, you should map a coordinate point to any other arbitrary coordinate point

As long as there exists an inverse transformation

ie you can map $x$ to $e^x$ if you so wish

hmm i guess

In vector spaces, those maps can only be linear

But there's no such restrictions in manifolds

it feels like every explanation of lorentz invariance/covariance and the lorentz transformations and rules uses different notation and different language, it's pretty upsetting

That's physics for you

:C

I try to keep the most general notation when I write physics, but it's pretty hard sometimes

Even there being a non-standard use of invariance and covariance happens

yeah

@Charlie We've talked about the two representations acting on the field operators before!

The statement of the question you link to there mentions the non-relativistic analogue which is useful to keep in the back of one's mind

Yeah I mean

It's also a thing just for rotations

$\hat{x}$ is such an operator

@Charlie Formally it all goes back to charts - a choice of coordinate system is a chart/set of charts, and a "coordinate transformation" between two coordinate systems is the concatenation of the map from one chart to the manifold and the map to the other chart from the manifold

It's not a field operator, but still

@ACuriousMind I thought the two vector spaces on which we have representations were the hilbert space of states and the vector spaces that the scalar, vector, tensor, spinor etc. objects live in

it was just surprising to me that we still talk about representations after second quantisation

Well you can still rotate, no matter how fancy your system gets!

but I guess it does make sense since the field operators act on the state space and so can be part of a representation

there's again an active/passive interpretation where you can choose to believe this concatenation is a diffeomorphism "acting on the manifold" (active) or that this is just a map between the coordinate systems (passive)

No matter how complicated your physics gets, you still have to get numbers in the end, and some of those numbers are positions

@Charlie that is correct, but neither of the two spaces stops existing after quantization (the first only comes into being there!)

And therefore depend on the coordinates you use

yeah

You can see it easily enough by using non-relativistic QM

Your state $\Psi$ has momentum $$\vec{p} \Psi(x) = -i\hbar \vec{\nabla} \Psi(x)$$

If you rotate your frame, the measured momentum is also rotated, $R(\vec{p})$

Therefore the wavefunction has to be changed so that its derivative is now $R(\vec{p})$

Wow! Gestapos are coming!

This time Indian ones!

@JohanLiebert Special Forces exist in every country! Bruh!

Special Forces can do arrests without warrant.

That's...not how the rule of law works.

Well, insofar as everyone can make provisional arrests without a warrant.

^That's the part, which Chinese Communist Party funded media didn't tell us.

@FadedGiant You're supposed to have some reason at least, in not-dictatorships

probable causes and all that

.The forces will later have to produce appropriate proof before the court.

What happens if they do not

Is the important part

No person can be arrested for more than 24 hours and must be produced before court in a limited time frame.

Policing the Police Laws are too there, I guess it should be applicable for Special Forces too.

Y'all still wear masks?

Masks are still mandatory in some countries

Masks are in vogue

One of the crazier things about exceptional Lie algebras is that although they all arise for Lie algebras over $\mathbb{C}$, when you go to fields like the Octonions they seem to arise 'naturally' (sort of i.e. it's still crazy) e.g. $G_2$ as the automorphism group of the octonions

"As such, the group $G_2$ is a higher analog of the symplectic group (which is the group that preserves a canonical 2-form on any $\mathbb{R}^{2n}$), obtained by passing from symplectic geometry to 2-plectic geometry."

@bolbteppa You can get all of them over $\mathbb{R}$ if you want (see e.g. mathoverflow.net/a/99795/157071). That you should get an exceptional group as the one that respects multiplication of the octonions doesn't strike me as that surprising - if you got one of the regular ones you'd be asking "Why is this one group in the sequence special?" and thinking that's "crazy"

Yeah, you get $SU(2)$ as the automorphism group for the quaternions, the other ones don't arise in this way though

$\mathbb{R}$ is the automorphism group on $\mathbb{R}$!

@bolbteppa it's obvious when you write it as SO(3) instead and remember that the multiplication of the quaternions is just the 3d cross product. (it's not obvious why you don't get SO(7) for the octonions, though - we don't have good intuition for the 7d cross product)

I have a question about crystal lattices. If I understand correctly, non-primitive lattices/unit cells are used because they capture the symmetry of the crystal, unlike the primitive lattices. Is this correct?

Yeah you do have to pass through $SO(7)$ to get to $G_2$ doing this, roughly the idea is you're trying to preserve norms i.e. $x_1^2+x_2^2+...$ so $SO(4)$ and $SO(8)$ arise for quaternions and octonions, transforming only the imaginary parts means reducing to $SO(3) = SU(2)$ and $SO(7)$ then [justification] you need to restrict to a subgroup we call $G_2$ for some reason hence $G_2$ is a subgroup of $SO(7)$

Isn't SU(2) the automorphism group for unit quaternions?

@abhas_RewCie Well! It seems you haven't lived in my state yet. I invite you to shift from your locality to some non-metropolitan city in UP and then feel the "rule of law". Here if you say against some high ranking official or someone related to them then, well, you have knocked at the doors of Death! And that used to happen in the absence of this law and I can see into the future what might happen now. So please don't just simply "bruh".

So you can at least build the generators of $G_2$ in terms of $SO(7)$ generators and have a plausible explanation for it, it's a real shame an easy picture like this doesn't arise for the others

Here's some evidence (if you want you can search for more):

google.com/amp/s/www.indiatoday.in/amp/india/story/…

$SO(7)$ has $6 \cdot 7 / 2 = 21$ generators, $G_2$ has $14$, 7 of the $SO(7)$ do something involving imaginary parts of the octonions, the remaining 14 are $G_2$...

Anybody know where I can nab a pdf copy of Ryders QFT book? My universities online library doesn't have it and I don't have 100 quid to shill out for it

Tried Library Genesis?

Oh do they have textbooks????

I thought that was just fiction books

Will have a check now

It has most textbooks

Beautiful

Thank you

They're gonna be getting a lot of traffic this year I imagine

I've always wondered what people spend their time putting books on libgen

I'm just glad they exist

These days a lot of books already exist in digital form

So it's probably fairly easy to put them there

This book is in .djvu so it might be that this is a scan

just converting it to pdf now

nope, normal pdf

I wonder why it was in .djvu. Probably storage reasons

What's the best Physics Paper you've ever seen/read in your entire life?

#SeriousQuestion #BestPaper #LifeQuestion

This isn't Twitter, keep your hashtags

@JakeRose It used to be that djvu provided better compression than pdf, and so it's still popular with "archive"-type sites

Idle question of no great significance: is a spinor always a spin half particle? If so then what is the corresponding object for a spin 3/2 particle like a gravitino?

@JohnRennie It's a "Rarita-Schwinger particle", see en.wikipedia.org/wiki/Rarita%E2%80%93Schwinger_equation

It's a fermion but I don't think many would call it a spinor

> a vector-valued spinor with additional components compared to the four component spinor in the Dirac equation

the funny thing is that it's not a spinor by Wiki's own definition via Clifford algebras :P

@ACuriousMind I tend to call it a Schwingor

What does the notation $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$ mean? The commutator of the two gamma matrices?

yes

@Slereah great, now I'm wondering what a Schwingor party is...

@JohnRennie The commutator of those components, yes

> The gravitino is naïvely taken to be a vector-spinor meaning that it has one index of each type, as in $\psi_{\mu \alpha}$. This has 16 components which are roughly speaking real or complex according to whether the field is 'Dirac' or 'Majorana' in a generalized sense (given simply by imposing the corresponding condition for each value of the spacetime index).

So a Rarita-Schwinger particle is a four component object like a spinor but each component is a four-vector.

A particle of spin $n/2$ can roughly be described as a field with $n$ spinor components

So for a spin $3/2$ particle that would be three spinor indices

Or one vector component and one spinor component

"It corresponds to the (1/2, 1/2) ⊗ ((1/2, 0) ⊕ (0, 1/2)) representation of the Lorentz group, or rather, its (1, 1/2) ⊕ (1/2, 1) part"

It's a beautiful autumn afternoon here in Chester. Sunshine, blue sky and around 22°C so nicely warm but not too hot. I am sitting gazing out of the window wondering if I should be doing something.

If it wasn't for global pandemics and our government about to make us a rogue state life would be idyllic.

Go get some scrumpy

All will be well

I'm pretty sure scrumpy will destroy COVID

'it's almost a cleaning'

Anytime I think I have free time I remember I don't even know why the Heterotic string can be related to $E_8$ or even what $E_8$ is really and I get back to work :'(

Symmetry group of the Leech lattice. So called because it sucks the life out of anyone who studies it.

Just second week of online classes and I’ve already had enough. I sorely miss board work and my books :(

Octonions 8 dimensional, the lattices are 8 dimensional, the 10D little group is 8 dimensional, it's all a big smorgasbord of 8's

The sign of Satan

888

Einstein turned Newton's 666 into the relativistic 888

Signs of alien life detected on Venus

@bolbteppa I was going to mention that when I got back from getting my coffee. That's really interesting. I'm not that into biology, so I don't really get the significance of what they found; but it definitely sounds pretty big, even if it somehow winds up not being life.

Yeah, I'm not sure if false positives are common with this stuff

I can think of at least two other occasions like that

When they found oxygen on a Jovian moon and when they found some bacteria on a meteor

So don't break out the party hats just yet

@JMac The explanation that I read is that "phosphorus and hydrogen hate each other," and so there's not a known pathway to continuously regenerate phosphine given the other chemicals in Venus's atmosphere.

Yeah the most disappointing result would be a bad measurement or false positive. It would be an interesting result even if we find it's from a chemical reaction though, because it's something that isn't expected apparently.

There's the famous line from Sherlock Holmes: "once you've eliminated the impossible, what remains, however improbable, must be the truth." But that mystery-novel phrasing doesn't allow enough for error or ignorance. The science version is, "once you've eliminated the impossible, you're about to learn something interesting."

@rob I had heard of that one before called out as the "Holmesian fallacy"

@JMac I like that name, but I also like (from the article) "appeal to omniscience."

Another version: "Nothing resembles a new effect quite so much as a mistake."

Yeah it's a good summary of the problem. I think the "Holmesian fallacy" is good for pointing out how dumb Sherlock's famous line was, but appeal to omniscience is a good way to see the general problem.

I always thought "elemtary my dear Watson" sounded patronizing

I never read Sherlock, but from the generic stuff I saw about it I always got the feeling he kinda was patronizing to Watson. No idea how the actual books handle it though.

iirc, Watson was an MD

I think that may be one of those things where the phrase was sensible, polite, and affectionate in the original, but has been used in a patronizing way in the popular culture over the 100+ years since. The past is a different country.

Ok, that makes sense

Some 100+ year stuff does stand the test of time and get labelled as "classic" though

@rob that sounds like a modern version :-)

Thanks for sharing.

@FadedGiant that's over 20 years of lost moderator experience including dmckee :(

over close to